In Exercises 5–8, use a significance level of A = 0.05 and refer to the

accompanying displays.

Casino Size and Revenue The New York Times published the sizes (square feet) and revenues (dollars) of seven different casinos in Atlantic City. Is there sufficient evidence to support the claim that there is a linear correlation between size and revenue? Do the results suggest that a casino can increase its revenue by expanding its size?

The variables under study are the sizes(square feet) and revenues (dollars) ofseven casinos. The correlation between the two variables is 0.44456896.

Let\(\rho \)be the true correlation coefficient measure for the two variables.

To test the claim that there is a linear correlation between two variables, form the hypotheses as shown below:

\(\begin{array}{l}{{\rm{H}}_{\rm{o}}}:\rho = 0\\{{\rm{{\rm H}}}_{\rm{a}}}:\rho \ne 0\end{array}\)

The samples number of casinos is7(n).

The test statistic is computed as follows:

\(\begin{aligned}{c}t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.44456896}}{{\sqrt {\frac{{1 - {{0.44456896}^2}}}{{7 - 2}}} }}\\ &= 1.110\end{aligned}\)

Thus, the test statistic is 1.110.

The degree of freedom is computedbelow:

\(\begin{aligned}{c}df &= n - 2\\ &= 7 - 2\\ &= 5\end{aligned}\)

The p-value is computed using the t-distribution table:

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T < t} \right)\\ &= 2P\left( {T < 1.110} \right)\\ &= 0.3175\end{aligned}\)

Thus, the p-value is 0.3175.

Since the p-value is greater than 0.05, the null hypothesis fails to be rejected.

Thus, it can be concluded at the 0.05 level of significance,there is not enough evidence to support theexistence of a linear correlation between size and revenue.

As there is no correlation between the variables, theyare not linearly associated.

Moreover, correlation does not implythat one variable causes the other to occur.